The mass of Al metal that releases 2.4 L of hydrogen gas at STP is calculated as below
AT STP 1 mole =22.4 l
what about 22.4l =? moles
(22.4 L x 1 mole) /22.4 L= 2mole
write the reacing reaction
2Al+ 6 HCL = 2AlCl3 + 3H2
by use of mole ratio between Al to H2 which is 2 : 3 the mole Al = 1 x2/3 = 0.667 moles
mass =moles x molar mass
=0.667 mole x27 g/mol= 18 grams
Answer:
- Mass of NaH₂PO₄·H₂O = 8.542 g
- Mass of Na₂HPO₄ = 5.410 g
Explanation:
Keeping in mind the equilibrium:
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺
We use the Henderson-Hasselbalch equation (H-H):
pH = pka + ![log\frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
For this problem [A⁻] = [HPO₄⁻²] and [HA] = [H₂PO₄⁻]
From literature we know that pka = 7.21, from the problem we know that pH=7.00 and that
[HPO₄⁻²] + [H₂PO₄⁻] = 0.100 M
From this equation we can <u>express [H₂PO₄⁻] in terms of [HPO₄⁻²]</u>:
[H₂PO₄⁻] = 0.100 M - [HPO₄⁻²]
And then replace [H₂PO₄⁻] in the H-H equation, <u>in order to calculate [HPO₄⁻²]</u>:
![7.00=7.21+log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]} \\-0.21=log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\10^{-0.21} =\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\0.616*(0.100M-[HPO4^{-2}])=[HPO4^{-2}]\\0.0616 M = 1.616*[HPO4^{-2}]\\0.03812 M =[HPO4^{-2}]](https://tex.z-dn.net/?f=7.00%3D7.21%2Blog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%5D%20%7D%7B0.100%20M-%5BHPO4%5E%7B-2%7D%5D%7D%20%5C%5C-0.21%3Dlog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%5D%20%7D%7B0.100%20M-%5BHPO4%5E%7B-2%7D%5D%7D%5C%5C10%5E%7B-0.21%7D%20%3D%5Cfrac%7B%5BHPO4%5E%7B-2%7D%5D%20%7D%7B0.100%20M-%5BHPO4%5E%7B-2%7D%5D%7D%5C%5C0.616%2A%280.100M-%5BHPO4%5E%7B-2%7D%5D%29%3D%5BHPO4%5E%7B-2%7D%5D%5C%5C0.0616%20M%20%3D%201.616%2A%5BHPO4%5E%7B-2%7D%5D%5C%5C0.03812%20M%20%3D%5BHPO4%5E%7B-2%7D%5D)
With the value of [H₂PO₄⁻],<u> we calculate [HPO₄⁻²]</u>:
[HPO₄⁻²] + 0.0381 M = 0.100 M
[HPO₄⁻²] = 0.0619 M
Finally, using the concentrations, the volume, and the molecular weights; we can calculate the weight of each substance:
- Mass of NaH₂PO₄·H₂O = 0.0619 M * 1 L * 138 g/mol = 8.542 g
- Mass of Na₂HPO₄ = 0.0381 M * 1 L * 142 g/mol = 5.410 g
Answer A,They can form a triple covalent bond.
Explanation:
Answer:
4.6 moles MgO
Explanation:
number of moles is equals to the mass in grams divided by RFM
RFM=16+24.3=40.3
187/40.3
=4.6 moles of MgO
Answer:
atmosphere
Explanation:
because oxygen is retained in the atmosphere